Quick answer Escape velocity follows from energy conservation as v_e = √(2GM/R) (G = 6.674×10⁻¹¹ N·m²/kg² = gravitational constant, M = body mass, R = body radius), proportional to √M and inversely to √R. For Earth it is about 11.2 km/s and for Jupiter 59.5 km/s.
Set mass and radius to calculate escape velocity for any celestial body
Parameters
Sliders are in log10. 0 = Earth value, 1 = 10× Earth, -1 = 1/10 Earth.
Results
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Escape velocity $v_e$
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Orbital velocity (first cosmic)
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Ratio $v_e / c$
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Schwarzschild radius
Escape velocity reaches or exceeds light speed at this radius — the body would collapse to a black hole.
Comparison with solar-system bodies
Escape velocity (km/s, log scale) for major bodies. The orange bar is the current configuration.
Escape velocity in plain language
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I keep hearing "escape velocity" — escape from what, exactly, and how fast does the rocket need to go?
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It's the minimum launch speed needed to leave a body's gravity well forever, without further thrust. From Earth that's about 11.2 km/s. Anything launched straight up at or above that speed coasts to infinity even with the engine off.
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11.2 km/s is roughly 135× a Shinkansen at 300 km/h. The simulator says Jupiter is 59.5 km/s — why is it so much higher?
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The formula is $v_e = \sqrt{2GM/R}$. Larger mass and smaller radius both raise it. Jupiter has 318× Earth's mass but 11× the radius, so the ratio is $\sqrt{318/11} \approx 5.4$ — close to the actual factor of 5.3. That's why launching from Jupiter would be impractical.
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When I pick the neutron-star preset, $v_e/c$ jumps above 0.6. So in a black hole the escape speed exceeds the speed of light?
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Right. The boundary is the Schwarzschild radius $r_s = 2GM/c^2$, where $v_e = c$. For the Sun, $r_s \approx 3$ km — compress the Sun into a 6-km sphere and it becomes a black hole. In nature this happens during gravitational collapse following a supernova.
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So the Moon has a much lower escape velocity. That's why the Apollo lunar module could lift off with a small engine?
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Exactly. The Moon's escape velocity is only 2.38 km/s, less than a fifth of Earth's. That's a key reason proposals for a lunar refueling base look attractive: launching deep-space missions from the Moon costs far less propellant than from Earth.
Deriving escape velocity
To carry an object from the surface of a body to infinity, its initial kinetic energy must equal the gravitational potential-energy difference. From conservation of mechanical energy:
where $G = 6.674 \times 10^{-11}$ N m² kg⁻² is Newton's gravitational constant, $M$ is the body's mass and $R$ its radius. Notably, $v_e$ does not depend on the projectile's mass $m$.
The orbital speed for a circular orbit grazing the surface ($GMm/R^2 = mv_o^2/R$) is:
So escape velocity is $\sqrt{2} \approx 1.414$ times the orbital velocity. Setting $v_e = c$ gives the Schwarzschild radius:
$$r_s = \frac{2GM}{c^2}$$
The same expression follows rigorously from general relativity (the Schwarzschild solution).
Escape velocities of major bodies
The bar chart spans many orders of magnitude: Moon (2.4 km/s), Earth (11.2 km/s), Jupiter (59.5 km/s), Sun (617 km/s). For a neutron star (1.4 solar masses, 10 km radius) the escape velocity reaches 60–70 % of the speed of light. Black holes are denser still and exceed it.
Implications for spaceflight: escape velocity differences directly drive mission cost. Going from Earth to Mars requires escaping Earth's well (11.2 km/s) and then a heliocentric delta-V to change orbit. A lunar refueling depot is attractive because the Moon's lower escape velocity makes deep-space launch much cheaper.
Common misconceptions
"Reaching escape velocity guarantees escape" is misleading: the formula assumes ballistic, engine-off flight. Real rockets keep thrusting, so they never need to reach $v_e$ instantaneously — they accelerate gradually and clear the atmosphere. The relevant figure of merit for mission planning is total delta-V, not $v_e$ alone.
Air resistance is ignored: the formula assumes vacuum. On Earth, atmospheric drag would dissipate huge energy from a hypothetical "space gun" that fires a projectile at $v_e$ at sea level. That's a major reason such systems aren't used in practice.
The minimum speed needed at a body's surface so that, with no further propulsion and ignoring air resistance, the projectile coasts to infinity instead of falling back.
Plug Earth's mass (5.97×10²⁴ kg) and radius (6371 km) into $v_e = \sqrt{2GM/R}$. The result is 11.186 km/s, conventionally rounded to 11.2 km/s.
The Moon has only 1/81 of Earth's mass and 27 % of its radius. The combination yields about 2.38 km/s, roughly one fifth of Earth's value.
At and inside the Schwarzschild radius the Newtonian formula gives $v_e \geq c$. The relativistic interpretation is that no signal — including light — can escape. The "escape speed" picture is a useful heuristic, but a full description requires general relativity.
Real-world use cases
Spaceflight engineering: propellant budgets for interplanetary missions are anchored to the launch body's escape velocity. Mission designers use the same $\sqrt{2GM/R}$ formula to size every transfer leg, plus delta-V margins for finite-thrust gravity losses.
Education and research: astronomy and astrophysics courses use the simulator to compare bodies that span six orders of magnitude in $v_e$, from asteroids to neutron stars.
CAE workflow integration: the escape velocity is a starting boundary condition for higher-fidelity tools — nozzle design (ANSYS, OpenFOAM), atmospheric re-entry thermal protection, structural sizing of upper stages — long before detailed FEM/CFD analysis begins.
Enter the celestial body's mass in kilograms using sMNum (e.g., Earth = 5.972×10^24 kg) and select the unit prefix with sM
Input the radius in meters via sRNum (e.g., Earth = 6.371×10^6 m) and confirm unit scale with sR
The simulator calculates escape velocity using v_e = √(2GM/r), orbital velocity as √(GM/r), the ratio to light speed, and Schwarzschild radius (2GM/c²)
Worked Example
For Jupiter: mass = 1.898×10^27 kg, radius = 7.149×10^7 m. Escape velocity = √(2 × 6.674×10^-11 × 1.898×10^27 / 7.149×10^7) = 59.5 km/s. Orbital velocity at surface = 42.1 km/s. Ratio v_e/c = 0.000198. Schwarzschild radius = 6.31 m (Jupiter remains a normal body, not a black hole).
Practical Notes
Black hole detection: When Schwarzschild radius exceeds actual radius, the object is a black hole; test with 10^30 kg in 10 m radius
Moon comparison: Luna (7.342×10^22 kg, 1.737×10^6 m) yields escape velocity 2.38 km/s versus Earth's 11.2 km/s
Accuracy depends on using standard gravitational constant G = 6.674×10^-11 m³/(kg·s²) for realistic astrophysical predictions
🎬 Watch it in motion
Orbital Mechanics | one speed bends a circle into a hyperbola #Shorts
